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Homework Help: Radius of convergence of a power series(complete result, hopefully)

  1. May 18, 2010 #1
    1. The problem statement, all variables and given/known data

    I have post this before but now I have come up with the complete result hopefully

    Anyway given the power series [tex]\sum_{j=0}^{\infty} F_{j} z^j [/tex]

    find the radius of convergence around zero and [tex]F_j = F_{j-1} - F_{j-2}[/tex]

    and that [tex]j \geq 2[/tex]

    3. The attempt at a solution

    I know that recursive relation [tex]F_j = F_{j-1} - F_{j-2}[/tex] can be rewritten to the polynomial [tex]z^ = z +1 [/tex] which has the roots

    [tex]z_{0},z_1 = \frac{1 \pm \sqrt{5}}{2}[/tex]

    and that I know from discrete mathematics that the general solution of [tex]F_j = A z_{0}^j + B z_{1}^j[/tex] and we know that if the series converges then

    [tex]\lim_{n \to \infty} |\frac{F_{j+1}}{F_j}| < 1[/tex]

    thus by the ratio test

    [tex]\lim_{n \to \infty} |\frac{F_{j+1}}{F_j}| = \lim_{n \to \infty} |\frac{\frac{\sqrt{5}+1}{2}^{n+1} A + \frac{\sqrt{5}+1}{2}^{n+1} B}{\frac{\sqrt{5}+1}{2}^{n} A+ \frac{\sqrt{5}+1}{2}^{n}B} | = z_0[/tex] and then

    [tex]\lim_{n \to \infty} |\frac{F_{j+1}z^{j+1}}{F_j z^{j}}| = |z|z_0 = \frac{1+\sqrt{5}}{2}|z| < 1[/tex] and hende the radius of Convergence of the series around zero is

    [tex] R = \frac{1+\sqrt{5}}{2}[/tex]

    How is that friends?

    Susanne
     
    Last edited: May 18, 2010
  2. jcsd
  3. May 18, 2010 #2

    vela

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    That's not correct. For example, if you had Fj=2j and z=1/4, you'd get

    [tex]\sum_{j=0}^\infty F_j z^j = 2^j 2^{-2j} = \sum_{j=0}^\infty 2^{-j} = 2[/tex]

    but |Fj+1/Fj| = 2 > 1. You can see that a series can converge even though the coefficients aren't bounded as long as zj goes to zero fast enough.

    Anyway, I think you're making this more complicated than it needs to be. If you plug in your solution for Fj into the original series, you get

    [tex]\sum_{j=0}^\infty F_j z^j = \sum_{j=0}^\infty (A z_0^j +B z_1^j) z^j = A \sum_{j=0}^\infty (z_0 z)^j + B \sum_{j=0}^\infty (z_1 z)^j[/tex]

    Can you see where to go from there?
     
  4. May 19, 2010 #3
    No I'm not sure please elaborate :)

    I can see what you are saying that I made an error so that the series diverges. So I need to show that both parts of the sum convergence?
     
  5. May 19, 2010 #4

    vela

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    Hint: The two series are geometric series.
     
  6. May 19, 2010 #5
    Vela,

    I have been looking in one of analysis books,

    If I view the series as [tex]\sum_{n=0}^{\infty} F_n(z-z_0)^n[/tex] and let

    [tex]\lim_{n \to \infty} sup |F_n|^{\frac{1}{n}}[/tex] and let [tex]r = \frac{1}{\lambda}[/tex]

    Where limit is [tex]+Inifity[/tex] and thusly by the theorem, r the radius of convergence is zero. Is this resonable argument here?
     
  7. May 19, 2010 #6
    And I know Vela,

    if I show that the two geometric series P1 and P2 converge on their own (Abel's theorem) then their sum

    P1 + P2 converge as well...

    I find [tex]A = \frac{1}{\sqrt{5}}[/tex] and [tex]B = -\frac{1}{\sqrt{5}}[/tex]

    and [tex]z_0, z_1 = \frac{1 \pm \sqrt{5}}{2}[/tex]

    But what I don't get Vela is that [tex]\lim{j \to \infty} \sum_{j = 0}^{\infty} (\frac{1 + \sqrt{5}}{2}z)^j [/tex]

    then for P1 being a geomtric series to converge if 0 < |z| < 1

    thereby [tex]P1 = \frac{1}{\sqrt{5}} \sum_{j = 0}^{\infty} (\frac{1 + \sqrt{5}}{2}z)^j = \frac{\frac{1 + \sqrt{5}}{2}}{1-z} = \frac{1+\sqrt{5}}{2(z-1)}[/tex]

    But what I don't get Vela is that [tex]\lim_{j \to \infty} ( (\frac{1 + \sqrt{5}}{2})^j|z| =0) [/tex]
     
    Last edited: May 19, 2010
  8. May 19, 2010 #7

    vela

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    You evaluated the limit incorrectly. Let [itex]z_\pm = (1\pm\sqrt{5})/2[/itex]. You have

    [tex]|F_n|^{\frac{1}{n}} = |Az_+^n+Bz_-^n|^{1/n} = z_+|A|^{1/n}\left|1+\frac{B}{A}\left(\frac{z_-}{z_+}\right)^n\right|^{1/n}[/tex]

    assuming A is not zero. You should be able to show that the lim sup is z+.
     
  9. May 19, 2010 #8
    okay don't I need to take the limit still?
     
  10. May 19, 2010 #9

    vela

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    Yes, and the sup as well. I was just rewriting |Fn|^(1/n) in a way to make it easier to analyze when determining the supremum and the subsequent limit of the sequence of sups.
     
  11. May 19, 2010 #10

    vela

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    I'm not sure what you did here. A and B are arbitrary constants. Unless you have additional info, like what F0 and F1 are, you can't determine them.
    To be honest, I have no idea what you're doing here. One thing I'll point out, though, is that the ratio of the geometric series isn't simply z.
     
  12. May 19, 2010 #11
    Maybe I am a bit behind but [tex]z_+[/tex] is larger than 1?
     
  13. May 19, 2010 #12
    I'm told that [tex]F_0 = F_1 = 1[/tex]
     
  14. May 19, 2010 #13

    vela

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    Yes, z+ is approximately 1.618 and z- is approximately -0.618. What's more important is that |z-/z+|<1.
     
  15. May 19, 2010 #14
    But radius of convergence which I'm suppose to end up with is still what some books call the golden ratio?
     
  16. May 19, 2010 #15

    vela

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    z+ is the golden ratio. The radius of convergence will be its reciprocal.
     
  17. May 19, 2010 #16
    Thanks now I know what I am suppose to end up with :)

    I will with the major hints you have given me re-do the calculations and then post them again then I get home for your (hopefully) approval :)
     
  18. May 20, 2010 #17

    Gib Z

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    I approached this differently and seemed to have got a result different to both of you. In particular, I find that the Radius of Convergence is 1, which is greater than the reciprocal of the Golden Ratio (~0.618).

    We start with the definition for the coefficients: [tex]F_0 = F_1 = 1, F_n = F_{n-1} - F_{n-2}[/tex]

    From this we have for n=0,1,2, F = 1, 1, 0 respectively.

    Also: [tex]F_n = F_{n-1} - F_{n-2} = ( F_{n-2} - F_{n-3} ) - F_{n-2} = - F_{n-3}[/tex]. So we can see the sequence for F is : 1 , 1, 0, -1, -1, 0 , 1, 1, 0, -1, -1, 0.....

    Hence our series certainly converges at least where [itex]\sum_{p=0}^{\infty} x^p [/itex] converges, that is, all x such that |x|< 1. But our series certainly doesn't converge if |x|= 1 or |x|>1, as the n-th term does not converge to zero.
     
  19. May 20, 2010 #18

    vela

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    D'oh! It appears both Susanne and I made the same mistake solving the recurrence relation for Fj. Using the correct solution, I find a radius of convergence of 1 as well.
     
    Last edited: May 20, 2010
  20. May 20, 2010 #19
    I get that solution to that version of the resursive relation to be -1? Is that correct then? This ruins everything :cry: now I get produce a general solution for F_j and I am totally lost on how you guys suddenly can conclude that the radius is insteed 1?

    By the way whats not clear to me how can suddenly out of the blue conclude you must rewrite the recursive relation to [tex]F_j= - F_{j-3}[/tex] where did that come from?
     
    Last edited: May 20, 2010
  21. May 20, 2010 #20

    Gib Z

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    I derived it in the original post. I wrote out the relation: F_n = F_(n-1) + F_(n-2)

    Then I replaced F_(n-1) by the equivalent expression given by the same rule: F_(n-1) = F_(n-2) - F_(n-3)

    Then I simplified. With that result, since we know the first 3 terms of the sequence, the rest come about easily and we can see why F is given by the sequence I gave in my previous post.
     
  22. May 20, 2010 #21
    Are you a 100% sure that is applicable here? Cause we are dealing with as I wrote j or n as you wrote which are larger than 2? and the original relation satisfies positive fibunacci numbers why re-write and deal with the negative once as well?

    I would very appriacate if you would elaborate on why this re-write is needed?

    and why is your approach then more correct than mine?
     
  23. May 20, 2010 #22

    Gib Z

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    The original relation does NOT satisfy the Fibonacci numbers, which require F_0 = F_1 = 1, and F_n = F_(n-1) + F_(n-2). For YOUR question, there is a minus sign, not plus.

    For our sequence F_n, which is not the Fibonacci sequence, we can show that F_n = -F_(n-3) if n is equal to or bigger than 3. So by the first post, we can derive the sequence F_n.

    That sequence allows us to know what our power series actually is, and eventually allows us to find the Radius of Convergence for it.
     
  24. May 20, 2010 #23
    Okay,

    Sorry I was a bit tired. If I change the original relation to F_n = F_(n-1) + F_(n-2) insteed of F_n = F_(n-1) + F_(n-2) will my original calculations on the radius of covergence fit? That the radius of convergence is the reciprocal of the golden ratio?
     
  25. May 20, 2010 #24

    Gib Z

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    Did you mean if you change the original relation to F_n = F_(n-1) + F_(n-2) instead of F_n = F_(n-1) - F_(n-2) ?

    If the coefficients of the power series were the Fibonacci numbers, then the radius of convergence would be the reciprocal of the golden ratio, yes.
     
  26. May 20, 2010 #25
    yes I mean that and it still satisfies that F(0) = F(1) = 1 and n>= 2....
     
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